A Dynamics Problem Regarding Friction.

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    Dynamics Friction
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Homework Help Overview

The discussion revolves around a dynamics problem involving friction between two masses, specifically focusing on the coefficients of static friction and the time it takes for the top mass to fall off the bottom mass after sliding begins.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the acceleration of the top mass and the time it takes to slide a distance of 1 meter. Some participants question the reference frame for acceleration and whether the acceleration calculated is appropriate for answering the posed question.

Discussion Status

Participants are actively engaging with the problem, clarifying the reference frames and the implications of the calculated acceleration. There is a recognition that the acceleration needs to be considered relative to the bottom block, indicating a productive direction in the discussion.

Contextual Notes

The original poster has provided specific values for the coefficients of friction and the masses involved, but there may be assumptions about the system's setup that are still under discussion.

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Homework Statement



SSRnLDS.png


That was the given information. Here are the questions they ask, I have solved a) and b), but I am completely stuck on c). I'll just put them here in case information from them are necessary.

a) What is the coefficient of static friction between the 5.0 kg mass and the horizontal surface? (answer 0.33)

b) What is the coefficient of static friction between the two masses?

c) From the point when the top mass starts sliding how long will it take for the top mass to fall off the bottom mass? (0.97s)

Homework Equations



F=ma
fric = μ|Fn|

The Attempt at a Solution



Block on Top (2.0kg)
ƩFx=-fk
ƩFx=-(0.3)(2kg*9.81m/s^2)
ma=-(0.3)(2kg*9.81m/s^2)
2kg*a=-5.886N
a= \frac{-5.886N}{2kg}
a= -2.943m/s^2

then...
Δd=viΔt+\frac{1}{2}aΔt^2
-1m=\frac{1}{2}2.943m/s^2*Δt^2
-1m=-1.4715m/s^2*Δt^2
-1m=-1.4715m/s^2*Δt^2
\frac{-2m}{-1.4715m/s^2}=Δt^2
Δt^2=0.68s^2
Δt=0.82s^Which is wrong =(
 
Last edited:
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You've found the 2kg's acceleration relative to the ground (except, not sure why you have it as negative; isn't positive to the right?). Is that the acceleration you need to answer the question?
 
So does that mean I need to find the acceleration relative to the bottom block?
 
AvocadosNumber said:
So does that mean I need to find the acceleration relative to the bottom block?
You want the time it takes the top block to move 1m, starting from 'rest'. 1m relative to what? "rest" relative to what?
 
Oh... I see now... relative to the bottom block.
 
Quite so.
 

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